# How to prove monotonicity of this function?

Let $$a>0$$. How to prove that the function: $$f(x)=\frac{a x-1}{\log(a x)}\cdot\frac{\log x}{x-1},$$ is monotonic (depending on $$a<1$$ or $$a>1$$). I know that we can calculate the derivative and determine its sign, but this needs much of calculation. I'm wondering if we can decide the monotonicity using a simple trick.

• You can compute a table where each rows represent the sub part of your expressions (there are 4 parts here) and each column represent the points where each sub part of your expression change sign (would be at 1/a, and 1 here), then you can look at the signs of the subparts on both side of those values and finally get the signs of your overall function on the entire domain by multiplying all the rows. If you get only pluses or minuses then your function is monotonous. Tell me if I'm not clear enough. Jan 25 '20 at 18:21
• @Jeanba Please demonstrate this for $f(x) = x^2$. Jan 25 '20 at 18:52
• @epsilon--delta: my bad, what I had in mind was the method using not the function but the derivative of the functio which is anyways not what the OP is asking, thanks for pointing this out! Jan 25 '20 at 20:22

Consider \begin{align} g&:\mathbb{R}\to \mathbb{R}& g(t)&= \ln \frac{\mathrm{e}^t-1}{t} \end{align} (where the value at $$0$$ is defined by continuity). Then from $$\ln f(x) = g(\ln x + \ln a) - g(\ln x)$$ we know that $$\tfrac{1}{\ln a} f$$ is monotone for all $$a\neq 1$$ iff $$g$$ is convex. But $$g$$'s second derivative $$g''(t)=\frac{\cosh t-1-\tfrac{t^2}{2}}{t^2(\cosh t -1)}$$ is nonnegative.
Admittedly, since $$g(t)=t + \sum_{n=1}^{\infty}\frac{B_n}{n}\frac{t^n}{n!}$$ is itself an important elementary function, using the second derivative to reduce the result to inequalities of other elementary functions seems unsatisfactory. Maybe there's a more intrinsic route?
• thank you but the function is increasing for $a>1$ and decreasing for $a<1$. Your answer doesn't depend on $a$? Jan 26 '20 at 21:29
• The second equation can be rewritten $$\frac{\ln f(x)}{\ln a}=\frac{g(\ln x+\ln a)-g(\ln x)}{\ln a}$$ so that in either case we are solely interested in the first divided difference of $g$ increasing, i.e., $g$ being convex. Jan 26 '20 at 21:47
• There is a typo in the denominator of the formula for $𝑔″(𝑡)$. The expression in the parenthesis should not be squared. Besides the claim "$𝑓$ is monotone iff $𝑔$ is convex" is generally not valid (for any fixed $𝑎$). The replacement of the word "iff" with "if" would be here appropriate.